What is the equation of the line normal to #f(x)=6x^3+2x # at #x=3#?

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Answer 1 · 2016-12-18T03:52:10

#y=-x/164+27555/164#

Given -

#y=6x^3+2x#

#dy/dx=18x^2+2#

Slope, exactly at #x=3#

At #x=3; 18(3)^2+2=164#

This is the slope of the tangent #m_1=164#.
Slope of the normal #m_2=-1/(m_1)=-1/164#

At #x=3; 6(3)^3+2(3)=162+6=168#

We have to find the equation of the line passing through the point #(3, 168)# with the slope #-1/164#

#mx+c=y#
#-1/164(3)+c=168#
#c=168+3/164=(27552+3)/164=27555/164#

#y=-x/164+27555/164#